What is the extraneous solution to these equations? $\dfrac{x^2 + x}{x - 9} = \dfrac{90}{x - 9}$
Answer: Multiply both sides by $x - 9$ $ \dfrac{x^2 + x}{x - 9} (x - 9) = \dfrac{90}{x - 9} (x - 9)$ $ x^2 + x = 90$ Subtract $90$ from both sides: $ x^2 + x - (90) = 90 - (90)$ $ x^2 + x - 90 = 0$ Factor the expression: $ (x - 9)(x + 10) = 0$ Therefore $x = 9$ or $x = -10$ At $x = 9$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 9$, it is an extraneous solution.